SAMPLE SPACES AND EVENTS

3.1. Let A and B be events. Find an expression and exhibit the Venn diagram for the event that: (i) A but not B occurs, i.e. only A occurs; (ii) either A or B, but not both, occurs, i.e. exactly one of the two events occurs.

(i) Since A but not B occurs, shade the area of A outside of B as in Figure (a)below. Note that Be, the complement of B, occurs since B does not occur; hence A and Bc occurs. In other words, the event is A nBc.

(ii) Since A or B but not both occurs, shade the area of A and B except where they intersect as in Figure (b) above. The event is equivalent to A but not B occurs or B but not A occurs.

Now, as in (i), A but not B is the event A nBc, and B but not A is the event B nAc. Thus the given event is (A nBc) U ( BnAc).

3.2. Let A , B and C be events. Find an expression and exhibit the Venn diagram for the event that (i) A and B but not C occurs, (ii) only A occurs.

(i) Since A and B but not C occurs, shade the intersection of A and B which lies outside of C, as in Figure (a)below. The event is A nB n^{C C .}

(ii) Since only A is to occur, shade the area of A which lies outside of B and of C, as in Figure (b) above. The event is A nBcn ^{C C .}

33. Let a coin and a die be tossed; let the sample space

S

consist of the twelve elements:

S

= {Hl, H2,H3, H4, H5, H6, Tl, T2,T3,T4,T5,T6}

(i)

Express explicitly the following events: A = {heads and an even number ap- pear}, B = {a prime number appears}, C = {tails and an odd number appear}.

(ii) Express explicitly the event that:

^(a)

A or B occurs, ( b ) B and C occurs,

(c)

only B occurs.

(iii) Which of the events A, B and C are mutually exclusive?

CHAP. 31 INTRODUCTION TO PROBABILITY 45 (i) To obtain A, choose those elements of S consisting of an H and an even number: A =

{H2, H4, H6).

To obtain B, choose those points in S consisting of a prime number: B = (H2, H3, H5, T2, T3, T5).

To obtain C, choose those points in S consisting of a T and an odd number: C = (Tl, T3, T5).

(ii) (a)A or B = A u B = {H2, H4, H6, H3, H5, T2, T3, T5) (b) B and C = B n C = { T 3 , T 5 }

(c) Choose those elements of B which do not lie in A or C: BnAcnCC = (H3, H5, T2).

(iii) A and C are mutually exclusive since A

n

C = @.

FINITE PROBABILITY SPACES

3.4.

Suppose a sample space

S

consists of

4

elements:

S

= {al,

a2, a3, a4>.

Which func- tion defines a probability space on S?

(i) P(a1)

=

4, P(a2) = Q, P(a3) = &, P(a4) = 3.

(ii) P(a1)

=

8, P(a2)=

$,

P(a3)

= -$,

P(a4)= +.

(iii) P(a1)= 4, P(a2)

=

&, P(a3) = Q, P(a4) = Q.

(iv) P(a1)

=

8, P(a2)= &, P(a3) = &, P(a4)=

0.

(i) Since the sum of the values on the sample points is greater than one,

8 +

^Q

+ & + ⁵

⁼

^a,

77

the function does not define a probability space on S.

(ii) Since P(a3)=

-&,

a negative number, the function does not define a probability space on S.

(iii) Since each value is nonnegative, and the sum of the values is one,

6 + & + ^Q + ^Q

^{= 1, the}

function does define a probability space on S.

(iv) The values are nonnegative and add up to one; hence the function does define a probability space on S .

3.5.

Let

S =

{al,

a2, a3, a4},

and let P be a probability function on

S.

(i) Find P(a1)if P(a2) = Q, P(a3)

=

9, P(a4)= Q.

(ii) Find P(a1)and P(a2)if P(a3) = P(a4) = & ^{and P(u1) =} 2P(a2).

(iii) Find P(a1) if P((a2, as}) = 8, P(

^(a2,a4})=

4 ^and P(a2)= Q.

(i) Let P(al)= p. Then for P to be a probability function, the sum of the probabilities on the sample points must be one: p

+ ^Q +

^Q

+ ^$

^{= 1 or p =}

5.

(ii) Let P(a2)= p, then P(al) = 2p. Hence 2 p + p +

&+ &

= 1 or p =

9.

Thus P(a2)=

Q

and P(a1) == Q.

(iii) Let P(aJ = p. P(a3) = P({az,a3N

-

P(a2) =

Q - Q

=

-g

W 4 ) = Wa2, a4N

-

^{P(a,) =}

+ ^{- Q}

⁼

^Q

Then p

+ Q +

^Q

^{+ Q}

^{= 1 or p =}

^9,

^{that is, P(al)= Q.}

3.6. A coin is weighted so that heads is twice as likely to appear as tails. Find P ( T ) and P(H).

Let P(T) = p; then P(H) = 2p. Now set the sum of the probabilities equal to one: p

+

^{2p = 1}

or p =

9.

Thus P(T) = p = Q and P(H)= 2 p =

6.

46 INTRODUCTION TO PROBABILITY [CHAP. 3

Two men,

ml

and

m2,

and three women,

WI,

w z and w3, are in a chess tournament.

Those of the same sex have equal probabilities of winning, but each man is twice as likely to win as any woman. (i) Find the probability that a woman wins the tournament. (ii) If ml and

W I

are married, find the probability that one of them wins the tournament.

Set P(wl)= p ; then P(w2)= P(w3)= p and P(ml) = P(m2)= 2p. Next set the sum of the probabilities of the five sample points equal to one: p

+

^p

+

^p

+

^{2 p}

+

^{2 p = 1 or p =}

3.

We seek (i) P({wl,w2,w3})and (ii) P({ml, wl}).Then by definition, P({Wl,W2,w3})= P(Wl)

+

P ( W 2 )

+

^{P(w3) =}

³ + ³ + ³

⁼

+

= P(m,)

+

^P(Wl)=

³ + ³

⁼

^$

3.8. Let a die be weighted so that the probability

of

a number appearing when the die is tossed is proportional to the given number (e.g.

6

has twice the probability of appearing as

3).

Let A = {even number}, B

=

{prime number}, C

=

(odd number).

(i) Describe the probability space, i.e. find the probability of each sample point.

(ii) Find P(A),P(B)and P(C).

(iii) Find the probability that:

(a)

an even or prime number occurs; ( 6 ) an odd prime number occurs; ( c ) A but not B occurs.

(i) Let P(l) = p . Then P ( 2 )= 2 p , P(3) = 3 p , P ( 4 ) = 4 p , P(5)= 5 p and P ( 6 ) = 6 p . Since the sum of the probabilities must be one, we obtain p

+

^{2 p}

+

^{3 p}

+

^{4 p}

+

^5p

+

^{6 p = 1 or p = 1/21. Thus}

P(1) =

$,

^{P ( 2 )=}

6,

^{P(3) =}

3,

^{P ( 4 ) =}

2,

^{P ( 5 )=}

&,

^{P ( 6 )=}

3

(ii) P ( A ) = P ( { 2 , 4 , 6 } ) =

3,

^{P(B) =} P({2,3,5}) =

E,

P(C) = P({1,3,5}) =

3.

(iii) (a)The event that an even or prime number occurs is A UB = { 2 , 4 , 6 , 3 , 5 } , or t h a t 1 does not occur. Thus P(A U B ) = 1

-

P(l) = h. 20

(b) The event that an odd prime number occurs is B n C = {3,5}. Thus P(BnC) = P ( { 3 , 5 H = 5. 8

(c) The event that A but not B occurs is A nBc = { 4 , 6 } . Hence P ( AnBc) = P ( { 4 , 6 } ) =

g.

FINITE EQUIPROBABLE SPACES

3.9. Determine the probability

p

of each event:

(i) an even number appears in the toss of a fair die;

(ii) a king appears in drawing

a

single card from an ordinary deck of 52 cards;

(iii) a t least one tail appears in the toss of three fair coins;

(iv)

a

white marble appears in drawing

a

single marble from an urn containing

4

white, 3 red and 5 blue marbles.

(i) The event can occur in three ways (a 2 , 4 or 6 ) out of 6 equally likely cases; hence p = ₌

f.

(iij There are 4 kings among the 52 cards; hence p = ⁴ = B. 1

(iii) If we consider the coins distinguished, then there are 8 equally likely cases: HHH,HHT, HTH,HTT,THH,THT,TTH,TTT.Only the first case is not favorable to the given event;

7

hence p = 5.

(iv) There are 4

+

³

+

^{5 =} 12 marbles, of which 4 are white; hence p = =

$.

CHAP. 31 INTRODUCTION TO PROBABILITY 47

3.10.

Two cards are drawn a t random from an ordinary deck of 52 cards. Find the proba-

bility p that (i) both are spades, (ii) one is a spade and one is

a

heart.

There are (",") = 1326 ways to draw 2 cards from 52 cards.

(i) There are

(123)

= 78 ways to draw 2 spades from 13 spades; hence

number of ways 2 spades can be dr;= - 78 - -1

= number of ways 2 cards can be dra.wn 1326 17

(ii) Since there are 13 spades and 13 hearts, there are 13.13 = 169 ways to draw a spade and a heart; herice p = = ₁₀₂ 2

-

3.11.

Three light bulbs are chosen a t random from 15 bulbs of which

5

are defective.

Find the probability p that (i) none is defective, (ii) exactly one is defective, (iii)

at

least one is defective.

There are

(135)

= 455 ways to choose 3 bulbs from the 15 bu.lbs.

(i) Since there are 15 ^-5 = 10 nondefective bulbs, there are

('i)

^{= 120} ways to choose 3 non- defective bulbs. Thus p =

3

⁼

%.

(ii) There are ⁵ defective bulbs and

( y )

⁼ 45 different paiirs of nondefective bulbs; hence there are 5 45 = 225 ways to choose 3 bulbs of which one is d.efective. Thus p = 455 ²²⁵ =

z.

⁴⁵

(iii) The event that at least one is defective is the complement of the event that none are defective which has, by (i),probability

E.

^{Hence p = 1-}

3

⁼

E.

3.12. _Two

cards are selected a t random from

10

cards numbered 1 to 10. Find the proba- bility p that the sum is odd if (i) the two cards are drawn together, (ii) the two cards

are

drawn one after the other without replacement, (iii) the two cards

are

drawn one after the other with replacement.

(i) There are

( y )

= 45 ways to select 2 cards out of 10. The sum is odd if one number is odd and the other is even. There are 5 even numbers and 5 odd numbers; hence there are 5 . 5 = 25 ways of choosing an even and an odd number. Thus p = =

i.

(ii) There are 10.9 = 90 ways to draw two cards one after the other without replacement.

5 = 25 There are 5 5 = 25 ways to draw an even number and then an odd number, and 5 ways to draw an odd number and then an even number; hence p =

9

=

2

⁼

:.

(iii) There are 10.10 = 100 ways to draw two cards one :after the other with replacement. As in (ii), there are 5 . 5 = 25 ways to draw an even number and then a n odd number, and

25+25 50

5.5 = 25 ways to draw an odd number and then an even number; hence p = 100= 100 = p ¹

3.13.

Six married couples are standing in

a

room.

(i)

If

2 people are chosen a t random, find the probability p that (a) they are married, ( b )one

is

male and one is female.

(ii)

If

4 people are chosen

at

random, find the probability p that (a)

2

married couples are chosen,

( b )

no married couple

is

among the

4,

(c) exactly one married couple is among the 4.

(iii)

If

the 12 people are divided into six pairs, find the probability p that (a) each pair is married, ( b )each pair contains a male and a female.

48 INTRODUCTION TO PROBABILITY [CHAP. 3

(i) There are

( y )

^{=66 ways} to choose 2 people from the 12 people.

(a) There are 6 married couples; hence p =

&

⁼

A.

(b) There are 6 ways to choose a male and 6 ways to choose a female; hence p =gs6.6 =

fi.

(ii) There are

(y)

^{=495 ways to choose 4} people from the 12 people.

15 1

(a) There are

(i)

⁼15 ways to choose 2 couples from the 6 couples; hence p =495 =33.

(b) The 4 persons come from 4 different couples. There are

t4)

6 =15 ways to choose 4 couples from the 6 couples, and there are 2 ways to choose one person from each couple. Hence

2.2-2.2.15

p = - -

49s -33'

(c) This event is mutually disjoint from the preceding two events (which are also mutually disjoint) and a t least one of these events must occur. Hence p

+ & +

^{=1 or p = I s33}

(iii) There are 2!21i:A!2!2! =26 12!ways to partition the 12 people into 6 ordered cells with 2 people in each.

(a) The 6 couples can be placed into the 6 ordered cells in 6! ways. Hence p =^12!/26 =^10,395 (b) The six men can be placed one each into the 6 cells in 6! ways, and the 6women can be

placed one each into the 6 cells in 6! ways. Hence p =8!6!^12!/26

- -

^{231 *}

3.14. A class contains 10 men and 20 women of which half the men and half the women have brown eyes. Find the probability

p

that a person chosen a t random is a man or has brown eyes.

Let A ={person is a man} and B ={person has brown eyes}. We seek P(A UB).

Then P ( A )= =f, _{P(B)= =}

i,

^{P ( AnB) =}

6

=

$.

Thus by Theorem 3.5, p = P ( A u B ) = P ( A )

+

^P(B)

^-

P ( A n B ) =

g + + ^{- Q}

^{= Q}